算术表达式为printf arg

Question

I am not able to understand why

float x = (4+2%-8);
printf("%f\n", x);

prints 6.000000 and

printf("%f\n", (4+2%-8));

prints 0.000000. Any information will be helpful. Regards.

Answer 1

The expression (4 + 2 % -8) produces an integer and you are trying to print a float (and they don't match).

In the first case the integer is converted to float (because of the assignment) so later on the printf works because the value is in a format %f expects.

Try this:

printf("%f\n", (4.0 + 2 % -8));
                 ^

Answer 2

It's because here:

float x = (4+2%-8);

The (4+2%-8) is of type int but is converted to float because that's the type of x . However, here:

printf("%f\n", (4+2%-8));

No cast is performed so you pass an int where it expects a float giving you a garbage value. You can fix this with a simple cast:

printf("%f\n", (float)(4+2%-8));

Answer 3

In the first snippet the resulting int value is implicitly converted to float due to assignment, in the second snippet you are lying to the compiler: you tell it to expect a value of type double but "send" a value of type int instead.

Do not lie to the compiler. It will get its revenge.

Note that the printf conversion specifier "%f" expects a value of type double , but "sending" a float is ok because that value is automagically converted to double before the function is called.